Lemma 68.18.6. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Let $y \in |Y|$ and assume that $y$ is represented by a quasi-compact monomorphism $\mathop{\mathrm{Spec}}(k) \to Y$. Then $|X_ k| \to |X|$ is a homeomorphism onto $f^{-1}(\{ y\} ) \subset |X|$ with induced topology.

** Fibers of field points of algebraic spaces have the expected Zariski topologies. **

**Proof.**
We will use Properties of Spaces, Lemma 66.16.7 and Morphisms of Spaces, Lemma 67.10.9 without further mention. Let $V \to Y$ be an étale morphism with $V$ affine such that there exists a $v \in V$ mapping to $y$. Since $\mathop{\mathrm{Spec}}(k) \to Y$ is quasi-compact there are a finite number of points of $V$ mapping to $y$ (Lemma 68.4.5). After shrinking $V$ we may assume $v$ is the only one. Choose a scheme $U$ and a surjective étale morphism $U \to X$. Consider the commutative diagram

Since $U_ v \to U_ V$ identifies $U_ v$ with a subset of $U_ V$ with the induced topology (Schemes, Lemma 26.18.5), and since $|U_ V| \to |X_ V|$ and $|U_ v| \to |X_ v|$ are surjective and open, we see that $|X_ v| \to |X_ V|$ is a homeomorphism onto its image (with induced topology). On the other hand, the inverse image of $f^{-1}(\{ y\} )$ under the open map $|X_ V| \to |X|$ is equal to $|X_ v|$. We conclude that $|X_ v| \to f^{-1}(\{ y\} )$ is open. The morphism $X_ v \to X$ factors through $X_ k$ and $|X_ k| \to |X|$ is injective with image $f^{-1}(\{ y\} )$ by Properties of Spaces, Lemma 66.4.3. Using $|X_ v| \to |X_ k| \to f^{-1}(\{ y\} )$ the lemma follows because $X_ v \to X_ k$ is surjective. $\square$

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